Evaluate the definite integral: int_0^{pi /2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $I = \int_0^{\pi /2} \log(	an x) \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^{\pi /2} \log(\

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $I = \int_0^{\pi /2} \log(	an x) \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^{\pi /2} \log(	an(\frac{\pi}{2} - x)) \, dx$ $I = \int_0^{\pi /2} \log(\cot x) \, dx$ $I = \int_0^{\pi /2} \log(\frac{1}{	an x}) \, dx$ $I = \int_0^{\pi /2} (\log 1 - \log(	an x)) \, dx$ Since $\log 1 = 0$,we have: $I = - \int_0^{\pi /2} \log(	an x) \, dx$ $I = -I$ $2I = 0 \implies I = 0$.

Evaluate the definite integral: $\int_0^{\pi /2} \log(\tan x) \, dx$.

Similar Questions

$\int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx = $

$\int_{\pi / 11}^{9 \pi / 22} \frac{d x}{1+\sqrt{\tan x}} = $

$\int_0^\pi x \sin x \, dx = $

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\tan^4 x} dx = $ . . . . . . .

$\int_{0}^{\pi /2} {\sin 2x \log \tan x \, dx}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module